The right polar factorization involves writing a matrix as a product of two other matrices, one which
preserves distances and the other which stretches and distorts. First here are some lemmas which
review and add to many of the topics discussed so far about adjoints and orthonormal sets and such
things. This is of fundamental significance in geometric measure theory and also in continuum
mechanics. Not surprisingly the stress should depend on the part which stretches and distorts. See
[58].
Lemma A.10.1Let A be a Hermitian matrix such that all its eigenvalues are nonnegative.Then there exists a Hermitian matrix A^{1∕2}such that A^{1∕2}has all nonnegative eigenvalues and
( )
A1 ∕2
^{2} = A.
Proof:Since A is Hermitian, there exists a diagonal matrix D having all real nonnegative
entries and a unitary matrix U such that A = U^{∗}DU. Then denote by D^{1∕2} the matrix which is
obtained by replacing each diagonal entry of D with its square root. Thus D^{1∕2}D^{1∕2} = D. Then
define
A1∕2 ≡ U ∗D1∕2U.
Then
( )
A1∕2 2 = U∗D1∕2U U∗D1∕2U = U ∗DU = A.
Since D^{1∕2} is real,
( ∗ 1∕2 )∗ ∗( 1∕2)∗ ∗∗ ∗ 1∕2
U D U = U D (U ) = U D U
so A^{1∕2} is Hermitian. ■
Next it is helpful to recall the Gram Schmidt algorithm and observe how if you start with orthonormal
vectors, the Gram Schmidt process does not change them. This was presented earlier, but here it is stated
again as a review.
Lemma A.10.2Suppose
{w1,⋅⋅⋅,wr,vr+1,⋅⋅⋅,vp}
is a linearly independent set of vectors suchthat
{w1, ⋅⋅⋅,wr}
is an orthonormal set of vectors. Then when the Gram Schmidt process is appliedto the vectors in the given order, it will not change any of the w_{1},
⋅⋅⋅
,w_{r}.
This lemma immediately implies the following lemma.
Lemma A.10.3Let V be a subspace of dimension p and let
{w1, ⋅⋅⋅,wr}
be an orthonormal set ofvectors in V . Then this orthonormal set of vectors may be extended to an orthonormal basis forV,
{w1,⋅⋅⋅,wr,yr+1,⋅⋅⋅,yp}
Proof: First extend the given linearly independent set
{w1,⋅⋅⋅,wr }
to a basis for V and then apply
the Gram Schmidt theorem to the resulting basis. Since
{w1,⋅⋅⋅,wr }
is orthonormal it follows from
Lemma 2.0.5 the result is of the desired form, an orthonormal basis extending
{w1,⋅⋅⋅,wr }
.
■
Here is another lemma about preserving distance.
Lemma A.10.4Suppose R is an m × n matrix with m ≥ n and R preserves distances. ThenR^{∗}R = I.
Proof:Since R preserves distances,
|Rx |
=
|x |
for every x. Therefore from the axioms of the dot
product,
= 0 for all x,y because the given x,y were arbitrary. Let y = R^{∗}Rx − x to conclude
that for all x,
∗
R Rx − x = 0
which says R^{∗}R = I since x is arbitrary. ■
With this preparation, here is the big theorem about the right polar factorization.
Theorem A.10.5Let F be an m×n matrix where m ≥ n. Then there exists a Hermitian n×n matrixU which has all nonnegative eigenvalues and an m × n matrix R which preserves distances and satisfiesR^{∗}R = I such that
F = RU.
Proof: Consider F^{∗}F. This is a Hermitian matrix because
∗ ∗
(F∗F) = F∗(F∗) = F ∗F
Also the eigenvalues of the n × n matrix F^{∗}F are all nonnegative. This is because if x is an
eigenvalue,
λ(x,x) = (F∗F x,x) = (Fx,F x) ≥ 0.
Therefore, by Lemma A.10.1, there exists an n×n Hermitian matrix U having all nonnegative eigenvalues
such that
2 ∗
U = F F.
Consider the subspace U
(Fn)
. Let
{Ux1,⋅⋅⋅,Uxr}
be an orthonormal basis for
n n
U (F ) ⊆ F .
Note that U
(Fn)
might not be all of F^{n}. Using Lemma A.10.3, extend to an orthonormal basis for all of
F^{n},
{U x1,⋅⋅⋅,U xr,yr+1,⋅⋅⋅,yn}.
Next observe that
{F x1,⋅⋅⋅,Fxr}
is also an orthonormal set of vectors in F^{m}. This is
because